SOLUTION: Can you please help me to sketch the graph, given is y=ax^2+bx+c where a<0 ; b<0 ; c<0 and ax^2+bx+c has only one solution.

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Question 1078174: Can you please help me to sketch the graph, given is y=ax^2+bx+c where a<0 ; b<0 ; c<0 and ax^2+bx+c has only one solution.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You mean that ax%5E2%2Bbx%2Bc=0 has only one solution.

If a<0 the graph opens down (like a frowny mouth),
and goes down without a bottom on the left end and on the right end,
like
graph%28200%2C200%2C-0.9%2C0.9%2C-0.19%2C0.1%2C-1.3%28x%2B0.2%29%5E2%29 , or graph%28200%2C200%2C-0.9%2C0.9%2C-0.18%2C0.2%2C-1.3%28x-0.2%29%5E2%2B0.1%29 , or graph%28200%2C200%2C-0.9%2C0.9%2C-0.19%2C0.1%2C-1.3%28x-0.2%29%5E2-0.1%29 .
That means there are at least some points below the x-axis.
In this specific case, because b<0 and c<0 too,
it is also obvious that for x=1 ax%5E2%2Bbx%2Bc=a%2Bb%2Bc%3C0 .
If there were also some points above the x-axis
the graph would have to cross the x-axis twice,
but the fact that ax%5E2%2Bbx%2Bc=0 has only one solution
means that it just touches it once and does not cross it,
like graph%28200%2C200%2C-0.9%2C0.9%2C-0.19%2C0.1%2C-1.3%28x%2B0.2%29%5E2%29 , or graph%28200%2C200%2C-0.9%2C0.9%2C-0.19%2C0.1%2C-1.1%28x-0.4%29%5E2%29 , or graph%28200%2C200%2C-0.9%2C0.9%2C-0.19%2C0.1%2C-x%5E2%29 .

I do not know if it was your teachers intention,
but the point where the graph touches the x- axis
must have x=-b%2F2a .
Did you learn in class that the vertex of the graph
is at x=-b%2F2a ?
If both (a and b) are negative,
that is x=-b%2F2a ,
so the graph can look like graph%28200%2C200%2C-0.9%2C0.9%2C-0.19%2C0.1%2C-1.3%28x%2B0.2%29%5E2%29 , or graph%28200%2C200%2C-0.9%2C0.9%2C-0.19%2C0.1%2C-1.1%28x%2B0.4%29%5E2%29 ,
touching the x-axis to the left of the origin.